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User blog:GamesFan2000/HNEAN (Part 3: Layered Arrays 1)
Oh, hello again! Remember the linear arrays of HNEAN? Good times those were! You still aren't impressed by the absolutely ridiculous growth we covered last time? Ok then, it's time to bring out the layered arrays! Grab your popcorn, relax, and enjoy! Introduction We left off last time having described how linear arrays work in HNEAN. Verdict on them: They are BIG! Too big for even a highly advanced alien society to express. But...there's still room to grow bigger. Lots of room. Now it's time to introduce you to the next tier of HNEAN, known as the layered arrays: {a{b}c} {3{1}3} will be our unlucky contestant! So, how does it work? Well, let me tell you: It creates a big crunch! Joking aside, it works like so: the first three is the entry value, and the second three is the length of the array. So, {3, 3, 3}. Oh, but there's a catch! The {1} means that you take the answer of the array you just created and create another array whose length and entry value are EQUAL to the answer of that first array! Yes, you just heard that correctly! Layered arrays completely destroy linear arrays right from the get-go. Oh, and you can chain these! In {3{1}3{1}3}, you solve from right to left, replacing everything after the first {1} with the answer to whatever everything after the first {1} was. And, arrays around the {1}'s are completely valid. So, what does {2} do? {3{2}3} I'll tell you what is does. It takes the chaining and the arraying to the extreme! And, we're not using {3, 3, 3} either. The array length is actually b^c! I made that generalized expression at the top for a reason! And whatever that (nine-entry array of threes)'s answer is determines how long the chain of {1}'s is and the length and entry value of the arrays between the {1}'s! As you can imagine, these are big decomposed arrays. For {3{3}3}, create a 27-entry array of threes. The answer of that determines the length of the chain of {2}'s and the length and entry value of the arrays around the {2}'s. Of course, we still have quite a way to go in the world of layered arrays. Arrays In The Layer Function Now we start putting arrays in those {#} separators. {3{1, 1}3} And what exactly do you think that equates to? Take the answer to {3, 3, 3} and create a chain such that the number between the separators is the aforementioned answer to the aforementioned array, the length of the chain is said answer, and the length and entry values of the arrays around the separators is also said answer. Now, if the second number in the separator array is 1, then the first entry has the same effect on length of the determiner array as it would if the first entry was alone. {3{1, 2}3} Now, if the second entry of the separator array is greater than one, we take the answer of the first entry exponentiated by the second surrounder number and tetrate it to the second entry in the separator. 3^^2 = 27, so it's a 27-length array of 3's. Take the answer of that and make it the first entry of our {n, 1} separators, the length of the chain of separators, and the length and entry value of the surrounding arrays. Pretty self-explanatory for how {n, n} separators work beyond {1, 2}. {a{b, c, d}e} arrays have you go ((b^e)^^c)^^^d to get the length of the determiner. For larger separator arrays, just add more unique hyper-operators. That is where I'll leave this notation for now. See you next time! Category:Blog posts